$g(x)=\begin{cases} e^x&\text{for }-5<x<-1 \\\\ \dfrac xe&\text{for }-1\leq x<0 \end{cases}$ Find $\lim_{x\to -1^-}g(x)$. Choose 1 answer: Choose 1 answer: (Choice A) A $-\dfrac1e$ (Choice B) B $0$ (Choice C) C $\dfrac{1}{e}$ (Choice D) D The limit doesn't exist.
Notice that we were asked to find the one-sided limit, $\lim_{x\to -1^-}g(x)$. This is the limit where $x$ -values approach $-1$ from the left. Let's find the limit as $x$ approaches $-1$ from the left. We will use the fact that $g(x)=e^x$ for $x$ -values smaller than $-1$. $\begin{aligned} &\phantom{=}\lim_{x\to -1^-}g(x) \\\\ &=\lim_{x\to -1^-}e^x \\\\ &=e^{-1}&\gray{\text{Direct substitution}} \\\\ &=\dfrac{1}{e} \end{aligned}$ In conclusion, we found that $\lim_{x\to -1^-}g(x)=\dfrac{1}{e}$.